Array Deletion at Middle Index in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to delete the element at the middle index of the array.

DSA C++

int mid = size / 2;
for (int i = mid; i < size - 1; i++) {
    arr[i] = arr[[1]];
}
size--;

Drag options to blanks, or click blank then click option'

Ai - 1

Bi + 1

Ci

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr[i - 1] instead of arr[i + 1] causes wrong shifting.

Not decreasing the size after deletion.

To delete the middle element, shift all elements after mid one position to the left by assigning arr[i] = arr[i + 1].

2fill in blank

medium

Complete the code to find the middle index correctly for odd-sized arrays.

DSA C++

int mid = (size - 1) [1] 2;

Drag options to blanks, or click blank then click option'

A/

B*

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division.

Not subtracting 1 before division.

The middle index for zero-based arrays is (size - 1) / 2 when size is odd.

3fill in blank

hard

Fix the error in the loop condition to avoid accessing out-of-bound elements during deletion.

DSA C++

for (int i = mid; i [1] size - 1; i++) {
    arr[i] = arr[i + 1];
}
size--;

Drag options to blanks, or click blank then click option'

A<

B>=

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes out-of-bound access.

Using '>' or '>=' causes the loop not to run.

The loop should run while i < size - 1 to avoid accessing arr[size], which is out of bounds.

4fill in blank

hard

Fill both blanks to correctly delete the middle element and update the array size.

DSA C++

int mid = size [1] 2;
for (int i = mid; i [2] size - 1; i++) {
    arr[i] = arr[i + 1];
}
size--;

Drag options to blanks, or click blank then click option'

A/

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for mid.

Using '>=' or '<=' in loop condition causing errors.

Divide size by 2 to find mid, and loop while i < size - 1 to shift elements correctly.

5fill in blank

hard

Fill all three blanks to create a function that deletes the middle element from an array.

DSA C++

void deleteMiddle(int arr[], int &size) {
    int mid = size [1] 2;
    for (int i = mid; i [2] size - 1; i++) {
        arr[i] = arr[[3]];
    }
    size--;
}

Drag options to blanks, or click blank then click option'

A/

B<

Ci + 1

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operators for mid calculation.

Incorrect loop condition causing out-of-bound errors.

Wrong element shifting causing data loss.

Divide size by 2 to find mid, loop while i < size - 1, and shift elements by assigning arr[i] = arr[i + 1].

Practice

1. What happens when you delete the middle element from an array of size 5?
int arr[5] = {10, 20, 30, 40, 50};
Delete the element at index 2 (middle). What is the new array?

easy

A. [20, 30, 40, 50]

B. [10, 20, 30, 40, 50]

C. [10, 20, 30, 50]

D. [10, 20, 40, 50]

Solution

1. Step 1: Identify the middle index and element

   The array size is 5, so middle index is 2. The element at index 2 is 30.

2. Step 2: Delete the middle element and shift remaining elements

   Remove 30, then shift elements after index 2 left by one: 40 moves to index 2, 50 moves to index 3.

3. Final Answer:

   [10, 20, 40, 50] -> Option D

4. Quick Check:

   Deleted middle element 30, shifted left [OK]

Hint: Remove middle, shift right elements left by one [OK]

Common Mistakes:

• Not shifting elements after deletion
• Deleting wrong index
• Leaving array size unchanged

2. Which of the following C++ code snippets correctly deletes the middle element from an array arr of size n?

easy

A. for(int i = n/2; i < n-1; i++) arr[i] = arr[i+1]; n--;

B. for(int i = 0; i < n/2; i++) arr[i] = arr[i+1]; n--;

C. arr[n/2] = 0; n--;

D. for(int i = n-1; i > n/2; i--) arr[i] = arr[i-1]; n--;

Solution

1. Step 1: Understand deletion logic

   To delete middle element at index n/2, shift all elements after it one position left.

2. Step 2: Check each code snippet

   for(int i = n/2; i < n-1; i++) arr[i] = arr[i+1]; n--; correctly shifts elements from middle to end left by one and reduces size. Others either shift wrong elements or just set zero without shifting.

3. Final Answer:

   for(int i = n/2; i < n-1; i++) arr[i] = arr[i+1]; n--; -> Option A

4. Quick Check:

   Shift elements after middle left [OK]

Hint: Shift elements from middle index to end left [OK]

Common Mistakes:

• Shifting elements before middle index
• Not reducing array size
• Assigning zero instead of shifting

3. What is the output of the following C++ code?

int arr[6] = {5, 10, 15, 20, 25, 30};
int n = 6;
int mid = n / 2;
for(int i = mid; i < n - 1; i++) {
  arr[i] = arr[i + 1];
}
n--;
for(int i = 0; i < n; i++) {
  std::cout << arr[i] << " ";
}

medium

A. 5 10 15 25 30

B. 5 10 15 20 25 30

C. 5 10 15 20 30

D. 5 10 20 25 30

Solution

1. Step 1: Calculate middle index and delete element

   n=6, mid=3. Element at index 3 is 20. We delete 20 by shifting elements after index 3 left.

2. Step 2: Shift elements and print array

   Shift arr[3]=arr[4]=25, arr[4]=arr[5]=30. New array: 5,10,15,25,30. Size reduced to 5.

3. Final Answer:

   5 10 15 25 30 -> Option A

4. Quick Check:

   Deleted 20, shifted left [OK]

Hint: Shift elements after middle left, reduce size by one [OK]

Common Mistakes:

• Not reducing array size
• Printing old array size
• Not shifting elements correctly

4. Find the error in this code that tries to delete the middle element from array arr of size n:

int mid = n / 2;
for(int i = mid; i <= n - 1; i++) {
  arr[i] = arr[i + 1];
}
n--;

medium

A. Array size n should not be decremented

B. Middle index calculation is wrong

C. Loop condition should be i < n - 1, not i <= n - 1

D. Loop should start from 0, not mid

Solution

1. Step 1: Analyze loop range for shifting elements

   The loop runs from mid to n-1 inclusive, so arr[i+1] accesses arr[n], which is out of bounds.

2. Step 2: Correct loop condition

   Loop should run while i < n-1 to avoid accessing beyond array size.

3. Final Answer:

   Loop condition should be i < n - 1, not i <= n - 1 -> Option C

4. Quick Check:

   Prevent out-of-bounds by correct loop limit [OK]

Hint: Loop till one before last index to avoid overflow [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Not decrementing array size
• Wrong middle index calculation

5. Given an array arr of size n, write the logic to delete the middle element and return the new array size. What will be the size after deletion if n is even?
Choose the correct approach.

hard

A. Delete element at index (n-1)/2, shift elements left, new size = n - 1

B. Delete element at index n/2, shift elements left, new size = n - 1

C. Delete element at index n/2, no shifting needed, new size = n - 1

D. Delete element at index n/2 - 1, shift elements left, new size = n - 1

Solution

1. Step 1: Understand middle index for even-sized array

   In standard array operations, middle index is n/2 (integer division), consistent with odd sizes.

2. Step 2: Delete element at n/2 and shift elements left

   Remove element at index n/2, shift all elements after it left by one, reduce size by 1.

3. Final Answer:

   Delete element at index n/2, shift elements left, new size = n - 1 -> Option B

4. Quick Check:

   Middle index = n/2 for even n [OK]

Hint: Use n/2 as middle index regardless of parity [OK]

Common Mistakes:

• Deleting lower middle index (n/2 - 1) for even n
• Not shifting elements after deletion
• Not reducing array size